Question

我在SQL Server 2008中有三个表，其设置如下：

员工表

empid(PK)
1
2

加入了EMPLOYEEATTRIBUTES

dataId(PK) | empId(FK) | attributeid | attributeVal
10 | 1 | A1 | somevalue1
20 | 1 | A2 | somevalue2
30 | 2 | A1 | somevalue3
40 | 2 | A3 | somevalue4

加入了ATTRIBUTES

attributeid | attributeName
A1 | attribute1
A2 | attribute2
A3 | attribute3

我需要将xml数据输出为以下格式

<rows>
   <row empid="1">
     <attribute1>somevalue1</attribute1>
     <attribute2>somevalue2</attribute1>
   </row>
   <row empid="2">
     <attribute1>somevalue3</attribute1>
     <attribute3>somevalue4</attribute1>
   </row>
</rows>

任何人都知道如何做到这一点??

Answer 1

如果您想跳过所有血腥细节并只看到答案，请查看此帖子底部的SQL查询。

这里的主要挑战是各种SQL Server FOR XML 选项无法生成所需输出中规定的动态元素名称。因此，我的第一个答案是考虑简单地返回传统的SQL结果集并让客户端生成XML。这是一个非常简单的流式转换。但是，这可能不是您的选择，因此我们继续让SQL Server生成XML。

我的第二个想法是使用SQL Server的内置XQuery功能来执行转换，因此：

/* WARNING: the following SQL does not work */
SELECT
  CAST((SELECT * FROM data FOR XML RAW) AS XML)
    .query('
      <rows>
        {
          for $empId in distinct-values(/row/@empId)
          return
            <row empid="{$empId}">
            {
              for $attr in /row[@empId = $empId]
              return
                attribute { "attribute" } { $attr/@attributeValue }
            }
            </row>
        }
      </rows>
    ')

唉，这不起作用。 SQL Server抱怨：

Msg 9315, Level 16, State 1, Line 25
XQuery [query()]: Only constant expressions are supported for the name expression
of computed element and attribute constructors.

显然，XQuery实现受到与 FOR XML 功能相同的限制。所以，我的第二个答案是建议在客户端生成XML :)但是如果你坚持从SQL生成XML，那么系好你的安全带......

总体战略是放弃SQL Server的本地SQL设施。相反，我们将使用字符串连接来构建XML文档。如果这种做法令人反感，你现在可以停止阅读：）

让我们从生成一个样本数据集开始：

SELECT NULL AS empId INTO employee WHERE 1=0
UNION SELECT 1
UNION SELECT 2

SELECT NULL AS dataId, NULL AS empId, NULL AS attributeId, NULL AS attributeVal INTO employeeAttributes WHERE 1=0
UNION SELECT 10, 1, 'A1', 'someValue1'
UNION SELECT 20, 1, 'A2', 'someValue2'
UNION SELECT 30, 2, 'A1', 'someValue3'
UNION SELECT 40, 2, 'A3', 'someValue4 & <>!'

SELECT NULL AS attributeId, NULL AS attributeName INTO attributes WHERE 1=0
UNION SELECT 'A1', 'attribute1'
UNION SELECT 'A2', 'attribute2'
UNION SELECT 'A3', 'attribute3'

请注意，我已经更改了提供示例中最后一个属性的值，以包含一些XML不友好的字符。

现在，将基本的SQL查询放在一起以执行必要的连接：

SELECT
  e.empId
, a.attributeName
, ea.attributeVal
FROM employee AS e
INNER JOIN employeeAttributes AS ea
  ON ea.empId = e.empId
INNER JOIN attributes AS a
  ON a.attributeId = ea.attributeId

给出了这个结果：

empId   attributeName   attributeVal
1       attribute1      someValue1
1       attribute2      someValue2
2       attribute1      someValue3
2       attribute3      someValue4 & <>!

最后一个属性中的那些有趣的角色会给我们带来麻烦。让我们改变查询以逃避它们。

; WITH
  cruftyData AS (
    SELECT
      e.empId
      , a.attributeName
      , (SELECT ea.attributeVal AS x FOR XML RAW) AS attributeValXml
      FROM employee AS e
      INNER JOIN employeeAttributes AS ea
        ON ea.empId = e.empId
      INNER JOIN attributes AS a
        ON a.attributeId = ea.attributeId
    )
, data AS (
    SELECT
      empId
    , attributeName
    , SUBSTRING(attributeValXml, 9, LEN(attributeValXml)-11) AS attributeVal
    FROM cruftyData
  )
SELECT * FROM data

结果：

empId   attributeName   attributeValXml
1       attribute1      someValue1
1       attribute2      someValue2
2       attribute1      someValue3
2       attribute3      someValue4 &amp; &lt;&gt;!

这可确保现在可以在XML文档中安全地使用属性值。属性名称怎么样？ XML属性名称的规则比元素内容的规则更严格。我们假设属性名称是有效的XML标识符。如果不是这样，则需要设计一些方案将数据库中的名称转换为有效的XML名称。这留给读者一个练习：）

下一个挑战是确保为每个员工将属性组合在一起，并且我们可以判断我们何时处于组中的第一个或最后一个值。这是更新的查询：

; WITH
  cruftyData AS (
    SELECT
      e.empId
      , a.attributeName
      , (SELECT ea.attributeVal AS x FOR XML RAW) AS attributeValXml
      FROM employee AS e
      INNER JOIN employeeAttributes AS ea
        ON ea.empId = e.empId
      INNER JOIN attributes AS a
        ON a.attributeId = ea.attributeId
    )
, data AS (
    SELECT
      empId
    , attributeName
    , SUBSTRING(attributeValXml, 9, LEN(attributeValXml)-11) AS attributeVal
    , ROW_NUMBER() OVER (PARTITION BY empId ORDER BY attributeName DESC) AS down
    , ROW_NUMBER() OVER (PARTITION BY empId ORDER BY attributeName) AS up
    FROM cruftyData
  )
SELECT * FROM data ORDER BY 1, 2

唯一的变化是将 down 和 up 列添加到结果集中：

empId  attributeName   attributeVal                down  up
1      attribute1      someValue1                  2     1
1      attribute2      someValue2                  1     2
2      attribute1      someValue3                  2     1
2      attribute3      someValue4 &amp; &lt;&gt;!  1     2

我们现在可以识别员工的第一个属性，因为 up 将 1 。可以使用 down 列以类似的方式识别最后一个属性。

有了这一切，我们现在有能力执行使用字符串连接构建XML结果的讨厌业务。

; WITH
  cruftyData AS (
    SELECT
      e.empId
      , a.attributeName
      , (SELECT ea.attributeVal AS x FOR XML RAW) AS attributeValXml
      FROM employee AS e
      INNER JOIN employeeAttributes AS ea
        ON ea.empId = e.empId
      INNER JOIN attributes AS a
        ON a.attributeId = ea.attributeId
    )
, data AS (
    SELECT
      empId
    , attributeName
    , SUBSTRING(attributeValXml, 9, LEN(attributeValXml)-11) AS attributeVal
    , ROW_NUMBER() OVER (PARTITION BY empId ORDER BY attributeName DESC) AS down
    , ROW_NUMBER() OVER (PARTITION BY empId ORDER BY attributeName) AS up
    FROM cruftyData
  )
, xmlData AS (
  SELECT
    empId
  , up
  , CASE WHEN up <> 1 THEN '' ELSE '<row id="' + CAST (empId AS NVARCHAR) + '">' END AS xml1
  , '<' + attributeName + '>' + attributeVal + '</' + attributeName + '>' AS xml2
  , CASE WHEN down <> 1 THEN '' ELSE '</row>' END AS xml3
  FROM data
)
SELECT xml1, xml2, xml3
--SELECT @result = @result + 'wombat' + xmlString
FROM xmlData
ORDER BY empId, up

结果：

xml1          xml2                                                 xml3
<row id="1">  <attribute1>someValue1</attribute1>        
              <attribute2>someValue2</attribute2>                  </row>
<row id="2">  <attribute1>someValue3</attribute1>        
              <attribute3>someValue4 &amp; &lt;&gt;!</attribute3>  </row>

剩下的就是将所有行连接在一起，并添加根行标记。由于T-SQL（尚未）具有字符串连接聚合函数，因此我们将使用变量作为累加器。 这是最后的查询，其所有的hacky荣耀：

DECLARE @result AS NVARCHAR(MAX)
SELECT @result = '<rows>'

; WITH
  cruftyData AS (
    SELECT
      e.empId
      , a.attributeName
      , (SELECT ea.attributeVal AS x FOR XML RAW) AS attributeValXml
      FROM employee AS e
      INNER JOIN employeeAttributes AS ea
        ON ea.empId = e.empId
      INNER JOIN attributes AS a
        ON a.attributeId = ea.attributeId
    )
, data AS (
    SELECT
      empId
    , attributeName
    , SUBSTRING(attributeValXml, 9, LEN(attributeValXml)-11) AS attributeVal
    , ROW_NUMBER() OVER (PARTITION BY empId ORDER BY attributeName DESC) AS down
    , ROW_NUMBER() OVER (PARTITION BY empId ORDER BY attributeName) AS up
    FROM cruftyData
  )
, xmlData AS (
  SELECT
    empId
  , up
  , CASE WHEN up <> 1 THEN '' ELSE '<row id="' + CAST (empId AS NVARCHAR) + '">' END AS xml1
  , '<' + attributeName + '>' + attributeVal + '</' + attributeName + '>' AS xml2
  , CASE WHEN down <> 1 THEN '' ELSE '</row>' END AS xml3
  FROM data
)
SELECT @result = @result + xml1 + xml2 + xml3
FROM xmlData
ORDER BY empId, up

SELECT @result = @result + '</rows>'
SELECT @result

XML最终出现在 @result 变量中。您可以使用以下方法检查它是否格式正确：

SELECT CAST(@result AS XML)

最终的XML如下所示：

<rows><row id="1"><attribute1>someValue1</attribute1><attribute2>someValue2</attribute2></row><row id="2"><attribute1>someValue3</attribute1><attribute3>someValue4 &amp; &lt;&gt;!</attribute3></row></rows>

Answer 2

你可以近距离接触 - 但你不能100％得到你想要的输出。

使用此查询：

SELECT
    EmpID AS '@empid',
    (
        SELECT 
           a.AttributeName AS '@name',
           ea.AttributeVal
        FROM dbo.EmployeeAttributes ea 
        INNER JOIN dbo.Attributes a ON ea.AttributeId = a.AttributeId
        WHERE ea.EmpID = e.EmpID
        FOR XML PATH ('attribute'), TYPE
    )
FROM dbo.Employee e
FOR XML PATH('row'), ROOT('rows')

你得到这个输出：

<rows>
  <row empid="1">
    <attribute name="Attribute1">
      <AttributeVal>SomeValue1</AttributeVal>
    </attribute>
    <attribute name="attribute2">
      <AttributeVal>SomeValue2</AttributeVal>
    </attribute>
  </row>
  <row empid="2">
    <attribute name="Attribute1">
      <AttributeVal>SomeValue3</AttributeVal>
    </attribute>
    <attribute name="attribute3">
      <AttributeVal>SomeValue4</AttributeVal>
    </attribute>
  </row>
</rows>

您不能做的是让内部XML节点具有与属性名称匹配的标记名称 - 您必须使用一些固定标记名称（如我的示例中的<attribute>），以及然后将从表中检索到的值作为这些XML标记的属性（如我的示例中的name=属性）或XML元素值应用。

据我所知，无法使用AttributeValue作为XML标记名称....

Answer 3

这是一个答案，但PIVOT命令限制了您必须提前知道属性的名称。稍微调整一下，您可以动态地执行此操作（尝试在SQL Server 2005中搜索动态数据透视）：

DECLARE @Employee TABLE ( empid INT )
DECLARE @EA TABLE
    (
      dataid INT
    , empid INT
    , attributeid CHAR(2)
    , AttributeVal VARCHAR(100)
    )
DECLARE @Attributes TABLE
    (
      AttributeID CHAR(2)
    , AttributeName VARCHAR(100)
    )

INSERT  INTO @Employee
VALUES  ( 1 ),
        ( 2 )

INSERT  INTO @EA
        ( dataid, empid, attributeid, AttributeVal )
VALUES  ( 10, 1, 'A1', 'somevalue1' )
    , ( 20, 1, 'A2', 'somevalue2' )
    , ( 30, 2, 'A1', 'somevalue3' )
    , ( 40, 2, 'A3', 'somevalue4' )

INSERT  INTO @Attributes
        ( AttributeID, AttributeName )
VALUES  ( 'A1', 'attribute1' )
        ,
        ( 'A2', 'attribute2' )
        ,
        ( 'A3', 'attribute3' )

SELECT  empID as '@empid'
      , attribute1
      , attribute2
      , attribute3
      , attribute4
FROM    ( SELECT    e.empid
                  , a.AttributeName
                  , ea.AttributeVal
          FROM      @Employee e
                    JOIN @EA ea ON e.empid = ea.empid
                    JOIN @Attributes a ON ea.attributeid = a.attributeid
        ) ps PIVOT
( MIN(AttributeVal) FOR AttributeName IN ( [attribute1], [attribute2], [attribute3], [attribute4] ) ) AS pvt    
FOR XML PATH('row'), ROOT('rows')

SQL Server从JOINed select语句生成XML数据行

3 个答案: